A poly-atomic molecule (Cv=3R, Cp=4R, where R is gas constant) goes from phase space point

Question

A poly-atomic molecule \(\left(\mathrm{C}_{\mathrm{v}}=3 \mathrm{R}, \mathrm{C}_{\mathrm{p}}=4 \mathrm{R}\right.,\) where R is gas constant) goes from phase space point \(\mathrm{A}\left(\mathrm{P}_{\mathrm{A}}=10^{5} \mathrm{~Pa}, \mathrm{~V}_{\mathrm{A}}=4 \times 10^{-6} \mathrm{~m}^{3}\right)\) to point \(\mathrm{B}\left(\mathrm{P}_{\mathrm{B}}=5\right. \left.\times 10^{4} \mathrm{~Pa}, \mathrm{~V}_{\mathrm{B}}=6 \times 10^{-6} \mathrm{~m}^{3}\right)\) to point \(\mathrm{C}\left(\mathrm{P}_{\mathrm{C}}=10^{4} \mathrm{~Pa}\right.,\) \(\mathrm{V}_{\mathrm{c}}=8 \times 10^{-6} \mathrm{~m}^{3}).\) A to B is an adiabatic path and B to C is an isothermal path.

The net heat absorbed per unit mole by the system is :

(1) \(500 \mathrm{R}(\text{l} n 3+\ln 4)\)

(2) \(450 \mathrm{R}(\text{l} n 4-\ln 3)\)

(3) \(500 \mathrm{R} \ \text{l} n 2\)

(4) \(400 \mathrm{R} \ \text{l} n 4\)

Answer 1

Correct option is  (2) \(450 \mathrm{R}(\ell n 4-\ell n 3)\)  

\(\Delta Q_{A B}=0\) adiabatic

\(\Delta \mathrm{Q}_{\mathrm{BC}}=\Delta \mathrm{W}_{\mathrm{BC}}\)

\(=n R T \ell\left(\frac{\mathrm{~V}_{\mathrm{C}}}{\mathrm{V}_{\mathrm{B}}}\right)=450 \mathrm{R} \ell \mathrm{n}\left(\frac{8 \times 10^{-6}}{6 \times 10^{-6}}\right)\)

\(=450 \mathrm{R} \ell n \left(\frac{4}{3}\right)=450 \mathrm{R}(\ell n 4-\ell n 3)\)

\(\therefore \Delta \mathrm{Q}=\Delta \mathrm{Q}_{\mathrm{AB}}+\Delta \mathrm{Q}_{\mathrm{BC}}\)

\(\Delta \mathrm{Q}=450 \mathrm{R}(\ell n 4-\ell n 3)\)

Note : Solution is based on direct data. B and C are not satisfying the condition of isothermal process.